[BZOJ2242][SDOI2011][BSGS][拓展欧几里得]计算器

题意

---

写一个计算器，能计算$A^B~mod~P$的值，$AX\equiv B(mod~P)$的解，和$A^X\equiv B(mod~P)$的解

---

数论杂题。
第一问快速幂，第二问exgcd，第三问BSGS.

#include <cstdio>
#include <map>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long ll;

int n,x,y,p,k,a,b;
map<int,int> Mp;

inline void reaD(int &x){
  char Ch=getchar();x=0;
  for(;Ch>'9'||Ch<'0';Ch=getchar());
  for(;Ch>='0'&&Ch<='9';x=x*10+Ch-'0',Ch=getchar());
}

inline ll solve_1(ll x,int y,int p){
  ll k=1;x%=p;
  while(y){
    if(y&1) k=1ll*k*x%p;
    x=1ll*x*x%p;
    y>>=1;
  }
  //printf("%lld\n",k);
  return k;
}

int gcd(int x,int y){return y?gcd(y,x%y):x;}

void exgcd(int x,int y,int &a,int &b){
  if(!y){a=1;b=0;return;}
  exgcd(y,x%y,a,b);
  int t=a;a=b;b=t-x/y*b;
}

inline void solve_2(int y,int z,int p){
  int g=gcd(y,p);
  if(z%g) {puts("Orz, I cannot find x!");return;}
  y/=g;z/=g;p/=g;
  exgcd(y,p,a,b);
  a=(1ll*a*z%p+p)%p;
  printf("%lld\n",a);
}

inline void solve_3(int y,int z,int p){
  y%=p;
  if(!y&&!z){puts("1");return;}
  if(!y){puts("Orz, I cannot find x!");return;}
  Mp.clear();
  ll t=ceil(sqrt(1.0*p)),x=1,ine=1;
  Mp[1]=0;
  for(int i=1;i<t;i++){
    x=1ll*x*y%p;
    if(Mp.count(x)) continue;
    Mp[x]=i;
  }
  ll k=solve_1(y,p-1-t,p);
  for(int i=0;i<t;i++){
    if(Mp.count(1ll*z*ine%p)){printf("%lld\n",Mp[1ll*z*ine%p]+1ll*i*t);return ;}
    ine=1ll*ine*k%p;
  }
  puts("Orz, I cannot find x!");
}

int main(){
  reaD(n);reaD(k);
  for(int i=1;i<=n;i++){
    reaD(x);reaD(y);reaD(p);
    if(k==1) printf("%lld\n",solve_1(x,y,p));
    else if(k==2) solve_2(x,y,p);
    else solve_3(x,y,p);
  }
}